# Phase Margin Question

Discussion in 'Homework Help' started by jegues, Feb 8, 2014.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
For convenience I've attached both the question and my attempt at the solution. Please see figures attached.

From the quadratic equation for the desired phase margin, I selected the w = 2.682 solution because I thought that we couldn't have negative values for w. (i.e. not physical)

However the correct solution comes about when w = +1.491, but the solution to my equation indicates w = -1.491.

How do we know when a negative frequency indicates an invalid solution, or a valid one where we can simply neglect the negative sign?

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
I believe it's a simple sign error in your equations:

You have

$-130^o=-90^o-atan(\frac{\omega}{4-\omega^2})$

which then becomes

$-40^o=-atan(\frac{\omega}{4-\omega^2})$

then

$40^o=atan(\frac{\omega}{4-\omega^2})$

then

$tan(40^o)=(\frac{\omega}{4-\omega^2})$

and so on ....